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  1. #1
    la vache fantôme phantomcow2's Avatar
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    Rudimentary question about finding upper/lower bound of integral

    Okay I am taking Calculus 2 this summer at my local university. It's a summer course so it's faster and therefore harder. I don't have a lot of access to help

    Use the min-max inequality to find the upper and lower bounds for the value of the integral from 0 to 1 of 1/(1-t^2)

    the lower bound can be found by f(minimum) * (b-a) and the upper by f(maximum) * (b-a)

    So when they say f(min) and max, what exactly do they mean? To find a function's minimum I would set the first derivative equal to zero. Do I plug this x value -- where the minimum occurs at -- or the y value, what the minimum value is, into the equation?
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  2. #2
    Senior Member MrCrassic's Avatar
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    I believe you input the minimum value that you found. I might be wrong; someone please verify.
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  3. #3
    la vache fantôme phantomcow2's Avatar
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    So by this you mean what the minimum/maximum is, the y value? I just want to be 100% clear. I think you're right.
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  4. #4
    L T X B O M P F A N S R apricissimus's Avatar
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    No, this is wrong! Plug in the x-value where the min/max occurs at.

    Suppose f(x) has a maximum on [a,b] at x_1 and a minimum at x_2.

    Then the upper bound your looking for would be f(x_1)*(b-a) and the lower bound would be f(x_2)*(b-a). This is equivalent to max*(b-a) and min*(b-a), respectively.

    Note that all you're doing is finding the smallest rectangle that entirely covers the actual area of the integral for the upper bound, and the largest rectangle that is entirely covered by the actual area of the integral for the lower bound.
    Last edited by apricissimus; 05-30-09 at 08:45 AM.

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